5. The decisive small-parameter calculation
The next phase no longer searched over unrelated formulas. We fixed
and asked whether this balanced power was always viable away from its zeros when \(k\) was sufficiently small.
5.1 The small-\(k\) first-order expansion
Let \(S=1-\nu_1^{2}-\cdots-\nu_\ell^{2}\). At a fixed point with \(S\gt{}0\), direct expansion of (3.3) and (3.6) gives
where
These were precisely the expressions implemented in the first-order numerical code: the input was \(P\), its first two derivatives, the point \(x\), and the frame coefficients \(\ip{e_i}{\tau_a}\).
Put
Substitution of (5.2) into the threshold polynomial (3.7) shows that
The ambient dimension \(n\) disappears from this first-order root; only the active rank \(\ell\) remains. Define the gap coefficient
Then
Consequently, the negative first-order term occurs in the normalized ratio \(s/W\), whereas the unnormalized threshold in (5.3) has the positive first-order expansion. The clean sufficient condition is
5.2 The first-order criteria in ranks two and three
For \(\ell=2\), equation (5.3) immediately gives
Equivalently, the first-order test is simply
For the balanced product (5.1), direct substitution and a Cauchy–Schwarz estimate give \(\varrho\gt{}0\) away from the zeros. Thus the balanced exponent \(2/m\) passes the rank-two first-order test.
For \(\ell=3\), put
The critical-root equation becomes
Accordingly, \(\varrho\gt{}0\) is equivalent to placing \(w\) to the right of the larger root. A convenient pair of scalar checks is
This was the form used in the rank-three tests.
5.3 Rank-three scans and analytic confirmation
The numerical work tested (5.6) for coordinate, symmetric, clustered, tilted, and random prescribed directions. Every entry in Table 5.1 concerns the first-order gap
For each listed configuration, we numerically evaluated the normalized boundary gap \(\varrho/P\) and the first-order gap \(\varrho\). The randomized rows record separate scans for \(m=2,\ldots,5\).
| Pole configuration | \(m\) | \(2/m\) | Boundary \(\min(\varrho/P)\) | Sampled \(\min\varrho\) |
|---|---|---|---|---|
| Coordinate triple | 3 | \(2/3\) | \(0.5006385\) | \(0.0999129\) |
| Tetrahedral four | 4 | \(1/2\) | \(0.5000182\) | \(0.3344756\) |
| Octahedral six | 6 | \(1/3\) | \(0.5000031\) | \(0.3157721\) |
| Random four | 4 | \(1/2\) | \(0.5009658\) | \(0.0426013\) |
| Random six | 6 | \(1/3\) | \(0.5000017\) | \(0.0427318\) |
| Random eight | 8 | \(1/4\) | \(0.5010803\) | \(0.1085162\) |
| Randomized two-pole (60 tries) | 2 | \(1\) | --- | \(2.41\times10^{-1}\) |
| Randomized three-pole (60 tries) | 3 | \(2/3\) | --- | \(2.06\times10^{-1}\) |
| Randomized four-pole (60 tries) | 4 | \(1/2\) | --- | \(2.28\times10^{-1}\) |
| Randomized five-pole (60 tries) | 5 | \(2/5\) | --- | \(2.92\times10^{-1}\) |
All four randomized families passed the first-order scans. The values in the table were numerical evidence only, but together they supported the stability of the balanced \(2/m\) product away from the zero set. The computation also showed why samples taken too close to a zero could be misleading at fixed \(k\): the first-order expansion is not uniform there.
After these scans, a detailed analytic calculation reduced (5.6) to a covariance estimate. It showed that \(w\) lies strictly to the right of the larger root in (5.5) for every set of distinct prescribed directions. Consequently, \(\varrho\) has a positive minimum on every compact set separated from the zeros. Continuity and (5.4) then give
there whenever \(k\gt{}0\) is chosen sufficiently small.
This completed the away-from-zero part for \(\ell\leq3\), but it did not solve the original problem. Near a prescribed direction,
and for every fixed positive \(k\) the uniform lower bound could still fail in a thin neighborhood of that zero. Decreasing \(k\) shrank the bad region but never removed it. The remaining task was therefore sharply defined: keep the successful \(2/m\) tail away from the zeros and replace its local vanishing model. This led to the linear calculation in the next section.